Arrange oxides $(M_{2} O_{3})$ of nitrogen family in increasing acidic nature:

S b_{2} O_{3} < A s_{2} O_{3} < B i_{2} S_{3} < P_{2} O_{3} < N_{2} O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

S b_{2} O_{3} > B i_{2} S_{3} > A s_{2} O_{3} > P_{2} O_{3} > N_{2} O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B i_{2} S_{3} < S b_{2} O_{3} < A s_{2} O_{3} < P_{2} O_{3} < N_{2} O_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B i_{2} S_{3} > S b_{2} O_{3} > A s_{2} O_{3} > P_{2} O_{3} > N_{2} O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $B i_{2} S_{3} < S b_{2} O_{3} < A s_{2} O_{3} < P_{2} O_{3} < N_{2} O_{3}$
As we move down the group, electronegativity decreases so acidic nature decreases and basic nature increases so order is:

$B i_{2} S_{3} < S b_{2} O_{3} < A s_{2} O_{3} < P_{2} O_{3} < N_{2} O_{3}$

Suggest Corrections

Similar questions

P_{2} O_{5}, A s_{2} O_{3}, S b_{2} O_{3}

B i_{2} O_{3}

P_{2} O_{3}

N_{2} O_{3}

N_{2} O_{3}

P_{2} O_{3}

K_{s p}

B i_{2} S_{3}

1.2 \times 10^{- 98}

B i_{2} S_{3}

S O_{2}, S i O_{2}, P_{2} O_{3}, A l_{2} O_{3}

Join BYJU'S Learning Program